[EM] General PR question (from Andy Jennings in 2011)
Kristofer Munsterhjelm
km_elmet at t-online.de
Wed Oct 1 11:33:56 PDT 2014
On 10/01/2014 04:16 PM, Kathy Dopp wrote:
> Tony,
>
> IMO, a better measure taking your approach, would be simply the sum of
> the measures of individual voters' representation and electing the set
> of candidates that maximizes this sum of representation measure. In
> fact, using such a summation measure would be, IMO, a good way to
> compare the satisfaction of voters with various voting methods.
>
> The problem with STV is that it would, in circumstances where there
> are three strong candidates, often result in much less satisfaction of
> voters according to the sum of individual voters' representation
> measure than approval voting and many other voting methods, due to the
> unfair way it treats different voters' ballots unequally, counting the
> 2nd and 3rd choices of some voters but not others due to its
> sequential consideration and elimination method.
What do you think of methods like Schulze STV that use a Condorcet-like
setup to consider all candidate combinations and thus avoid the path
dependence of the eliminations?
> if there were 4 seats to be elected, each voter's representation for
> set ABCE would be:
>
> 10 A 1/4
> 10 B 1/4
> 10 C 1/4
> 9 D 0
> 1 D,E 1/4
>
> Summing these measures for each voter would give 31/4
>
> If set ABCD were elected instead, the sum would be 39/4 because more
> voters are have contributed towards electing a representative. Thus,
> the set ABCD is preferable.
>
> Thus, I would gauge the representation for each voter using the
> formula n_i/s where n_i is the number of winning candidates each voter
> voted for. This seems like a more straight forward, simpler, still
> effective measure.
How about this?
51: ABCD
49: EFGH
Four to elect.
Consider first a proportional outcome, ABEF. This would give:
51: ABCD 2/4
49: EFGH 2/4
= 50.
Then consider the extremely majoritarian outcome ABCD:
51: ABCD 4/4
49: EFGH 0
= 51, so this has a greater score. But that's not proportional!
Perhaps I got your method wrong, though, but it seems you'd need some
sort of deweighting so that getting two candidates you approve of is
worth less than twice getting one candidate you approve of. Otherwise a
coordinated majority can fill the council completely.
PAV (proportional approval voting) is an example of such a rule: it
gives a voter 1 point for the first candidate both on the council and
the voter's approval ballot, 1/2 additional points for the second, 1/3
for the third and so on (reducing to D'Hondt in a party list situation).
Then it chooses winners so that the sum of points is maximized. There's
also birational voting, which is similar but with a different penalty
function; if I recall correctly, birational did better than PAV
according to my tests (
https://www.mail-archive.com/election-methods@lists.electorama.com/msg05937.html
).
It would be an interesting mathematical puzzle to determine such a
penalty function so that various desirable properties are satisfied.
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